poj1330 Nearest Common Ancestors

Posted 2020-08-29 王宜鸣

题意：

LCA裸题。

思路：

1. 朴素

2. 基于二分

3. 基于RMQ

实现：

1.

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <vector>
 4 #include <cstring>
 5 using namespace std;
 6 
 7 vector<int> G[10005];
 8 int t, n, x, y;
 9 int in[10005], parent[10005], depth[10005];
10 
11 void dfs(int v, int p, int d)
12 {
13     parent[v] = p;
14     depth[v] = d;
15     for (int i = 0; i < G[v].size(); i++)
16     {
17         if (G[v][i] != p)
18             dfs(G[v][i], v, d + 1);
19     }
20 }
21 
22 void init(int root)
23 {
24     dfs(root, -1, 0);
25 }
26 
27 int lca(int root, int x, int y)
28 {
29     while (depth[x] > depth[y])
30     {
31         x = parent[x];
32     }
33     while (depth[y] > depth[x])
34     {
35         y = parent[y];
36     }
37     while (x != y)
38     {
39         x = parent[x];
40         y = parent[y];
41     }
42     return x;
43 }
44 
45 int main()
46 {
47     cin >> t;
48     while (t--)
49     {
50         cin >> n;
51         for (int i = 1; i <= n; i++)
52         {
53             G[i].clear();
54         }
55         memset(depth, 0, sizeof(depth));
56         memset(in, 0, sizeof(depth));
57         memset(parent, 0, sizeof(parent));
58         for (int i = 0; i < n - 1; i++)
59         {
60             scanf("%d %d", &x, &y);
61             G[x].push_back(y);
62             in[y]++;
63         }
64         int root = 0;
65         for (int i = 1; i <= n; i++)
66         {
67             if (!in[i])
68             {
69                 root = i;
70                 break;
71             }
72         }
73         cin >> x >> y;
74         init(root);
75         cout << lca(root, x, y) << endl;
76     }
77     return 0;
78 }

2.

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <vector>
  4 #include <cstring>
  5 using namespace std;
  6 
  7 const int MAX_LOG_N = 14;
  8 
  9 vector<int> G[10005];
 10 int t, n, x, y;
 11 int in[10005], parent[MAX_LOG_N][10005], depth[10005];
 12 
 13 void dfs(int v, int p, int d)
 14 {
 15     parent[0][v] = p;
 16     depth[v] = d;
 17     for (int i = 0; i < G[v].size(); i++)
 18     {
 19         if (G[v][i] != p)
 20             dfs(G[v][i], v, d + 1);
 21     }
 22 }
 23 
 24 void init(int root)
 25 {
 26     dfs(root, -1, 0);
 27     for (int k = 0; k < MAX_LOG_N - 1; k++)
 28     {
 29         for (int i = 1; i <= n; i++)
 30         {
 31             if (parent[k][i] < 0)
 32             {
 33                 parent[k + 1][i] = -1;
 34             }
 35             else
 36             {
 37                 parent[k + 1][i] = parent[k][parent[k][i]];
 38             }
 39         }
 40     }
 41 }
 42 
 43 int lca(int root, int x, int y)
 44 {
 45     if (depth[x] > depth[y])
 46         swap(x, y);
 47     for (int k = 0; k < MAX_LOG_N; k++)
 48     {
 49         if ((depth[y] - depth[x]) >> k & 1)
 50         {
 51             y = parent[k][y];
 52         }
 53     }
 54     if (x == y)
 55     {
 56         return x;
 57     }
 58     for (int k = MAX_LOG_N - 1; k >= 0; k--)
 59     {
 60         if (parent[k][x] != parent[k][y])
 61         {
 62             x = parent[k][x];
 63             y = parent[k][y];
 64         }
 65     }
 66     return parent[0][y];
 67 }
 68 
 69 int main()
 70 {
 71     cin >> t;
 72     while (t--)
 73     {
 74         cin >> n;
 75         for (int i = 1; i <= n; i++)
 76         {
 77             G[i].clear();
 78         }
 79         memset(depth, 0, sizeof(depth));
 80         memset(in, 0, sizeof(depth));
 81         memset(parent, 0, sizeof(parent));
 82         for (int i = 0; i < n - 1; i++)
 83         {
 84             scanf("%d %d", &x, &y);
 85             G[x].push_back(y);
 86             in[y]++;
 87         }
 88         int root = 0;
 89         for (int i = 1; i <= n; i++)
 90         {
 91             if (!in[i])
 92             {
 93                 root = i;
 94                 break;
 95             }
 96         }
 97         cin >> x >> y;
 98         init(root);
 99         cout << lca(root, x, y) << endl;
100     }
101     return 0;
102 }

3.

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <vector>
  4 #include <cstring>
  5 #include <algorithm>
  6 using namespace std;
  7 
  8 const int MAXN = 10005;
  9 vector<int> G[MAXN];
 10 int t, n, x, y;
 11 int in[MAXN], depth[2 * MAXN - 1], id[MAXN], vs[2 * MAXN - 1];
 12 
 13 const int MAXM = 32768;
 14 const int INF = 0x3f3f3f3f;
 15 struct node
 16 {
 17     int index, d;
 18 };
 19 node data[2 * MAXM - 1];
 20 
 21 void update(int k, int a)
 22 {
 23     int tmp = k;
 24     k += n - 1;
 25     data[k].index = tmp;
 26     data[k].d = a;
 27     while (k)
 28     {
 29         k = (k - 1) / 2;
 30         if (data[2 * k + 1].d < data[2 * k + 2].d)
 31         {
 32             data[k].d = data[2 * k + 1].d;
 33             data[k].index = data[2 * k + 1].index;
 34         }
 35         else
 36         {
 37             data[k].d = data[2 * k + 2].d;
 38             data[k].index = data[2 * k + 2].index;
 39         }
 40     }
 41 }
 42 
 43 void rmq_init(int * depth, int k)
 44 {
 45     n = 1;
 46     while (n < k) n <<= 1;
 47     for (int i = 0; i < 2 * n; i++)
 48     {
 49         data[i].d = INF;
 50     }
 51     for (int i = 0; i < k; i++)
 52     {
 53         update(i, *(depth + i));
 54     }
 55 }
 56 
 57 node query(int a, int b, int k, int l, int r)
 58 {
 59     if (b <= l || a >= r)
 60     {
 61         node res;
 62         res.index = -1;
 63         res.d = INF;
 64         return res;
 65     }
 66     if (l >= a && r <= b)
 67         return data[k];
 68     node x = query(a, b, 2 * k + 1, l, (l + r) / 2);
 69     node y = query(a, b, 2 * k + 2, (l + r) / 2, r);
 70     if (x.d < y.d)
 71         return x;
 72     return y;
 73 }
 74 
 75 void dfs(int v, int p, int d, int & k)
 76 {
 77     id[v] = k;
 78     vs[k] = v;
 79     depth[k++] = d;
 80     for (int i = 0; i < G[v].size(); i++)
 81     {
 82         if (G[v][i] != p)
 83         {
 84             dfs(G[v][i], v, d + 1, k);
 85             vs[k] = v;
 86             depth[k++] = d;
 87         }
 88     }
 89 }
 90 
 91 int init(int root)
 92 {
 93     int k = 0;
 94     dfs(root, -1, 0, k);
 95     rmq_init(depth, k);
 96     return k;
 97 }
 98 
 99 int lca(int root, int x, int y)
100 {
101     node res = query(min(id[x], id[y]), max(id[x], id[y]) + 1, 0, 0, n);
102     return vs[res.index];
103 }
104 
105 int main()
106 {
107     cin >> t;
108     while (t--)
109     {
110         cin >> n;
111         for (int i = 1; i <= n; i++)
112         {
113             G[i].clear();
114         }
115         memset(depth, 0x3f, sizeof(depth));
116         memset(in, 0, sizeof(depth));
117         memset(id, 0, sizeof(id));
118         memset(vs, 0, sizeof(vs));
119         for (int i = 0; i < n - 1; i++)
120         {
121             scanf("%d %d", &x, &y);
122             G[x].push_back(y);
123             in[y]++;
124         }
125         int root = 0;
126         for (int i = 1; i <= n; i++)
127         {
128             if (!in[i])
129             {
130                 root = i;
131                 break;
132             }
133         }
134         int res = init(root);
135         cin >> x >> y;
136         cout << lca(root, x, y) << endl;
137     }
138     return 0;
139 }

总结：

还可以使用tarjan算法。